Objectives of the materials covered:
1.
The student should be able use the equations of statics to generate an influence line for the reaction, internal shear, and internal moment at any point in a simply supported beam.
Generation of influence lines: Probably the most straight-forward way to generate an influence line is to use statics. The process involves
1) determining what influence line is desired, i.e. for one of the reactions, for an internal axial force, for an internal shear, or for an internal moment,
2) placing a unit load at various locations along the structure,
3) use statics to calculate the resulting reaction, axial force, shear, or moment corresponding to that position of the unit load, and
4) plot the resulting magnitudes on the beam at the current position of the unit load.
Influence lines for reactions for a simply supported beam: As an example, assume that the influence line for the left reaction of a simply-supported beam is to be generated.
Place a unit load at the left end of the beam and solve for the magnitude of the resulting left reaction.
Figure 1
By summing moments about the right reaction, the resulting magnitude for the left reaction is found to be 1.0. This value of 1.0 is then plotted on the beam at the current point of application of the load, namely over the left reaction (see Figure 4.)
Next, move the unit load to the center of the beam and again solve for the left reaction.
By summing moments about the right reaction, the resulting magnitude for the left reaction is found to be 0.5. Plot this result of 0.5 at the current location of the unit load, i.e. at the center of the beam (see Figure 4 for plotting of all results.)

Figure 2
Finally, move the unit load to the right end of the beam and by statics solve for the left reaction. For this case, Ra = 0.0. Again, plot this result at the current location of the unit load, i.e. at the right end of the beam (see Figure 4.)
Figure 3
The result plot of these values is shown in Figure 4, and is simply a plot of the resulting magnitude of the left reaction corresponding to any position of the unit load. x
Figure 4
The result is an influence line for the left reaction of a simply supported beam. Although we only found values at three points along the beam, we can connect these points to produce a continuous curve. Additional intermediate values could be found if desired, but will be found to lie on the straight line plotted in Figure 4.

Influence lines for internal shear in a simply supported beam: Now assume that we desire to generate the influence line for shear at the quarter point in the same beam. It is found in the same manner – namely, a unit load is moved across the entire length of the beam, and the resulting shear at the quarter point is computed and plotted. The process is detailed below:
1) Position the unit load at the left reaction and solve for the resulting magnitude of the left reaction (see Figure 1 for result.)
2) Cut the beam of Figure 1 at the quarter point, where the shear V is desired, and draw a free body to expose the internal shear and moment at the quarter point in the beam.
V= 0
M= 0
Figure 5
3) Solve for the exposed shear V using ∑ Fv = 0. In this case we find that a unit load placed at the left end of the beam causes zero shear at the quarter point (see
Figure 5.)
4) Now move the unit load just slightly to the left of the quarter point on the beam and solve for the resulting reactions (see Figure 6.) The left reaction is found to be
0.75 by summing moments about the right reaction.
Figure 6
5) Again, draw a free-body of the left quarter of the beam, and solve for the exposed shear at the quarter-point by summing forces vertically (see Figure 7.) In this case, a resulting shear of 0.25 is caused by a unit load placed just to the left of the

quarter-point. The direction of shear shown in Figure 7 will arbitrarily be selected as negative.
V= 0.25
M= 0.1875L
Figure 7
6) At this time it should be emphasized that even though the load is moving across the structure, the free-body again, and always, must cut the beam at the quarter point of the beam, thus exposing the shear to be studied at that point.
7) Now move the unit load just slightly to the right of the quarter point on the beam and solve for the resulting reactions (again see Figure 6.) Since the load is moved only an infinitesimal distance to the right of its previous position, the reaction remains = 0.75. However, even though this movement does not change the reaction, it now moves the unit load outside of the free-body used to solve for the resulting shear at the quarter point. Thus it changes the resulting shear to a positive 0.75 (see Figure 8.) Again note that the free-body remains L/4 long for this, and for all other studies when determining the influence line for shear at L/4.
V= 0.75
M= 0.1875L
Figure 8
8) Next, move the unit load to the center of the beam and solve for the left reaction
(see Figure 2.)
9) Again, draw a free-body of the left ¼ beam, and solve for the exposed shear by summing forces vertically. In this case, a resulting shear of positive 0.5 is caused when the unit load is placed at the center of the beam (see Figure 9.)

V= 0.50
M= 0.125L
Figure 9
10) Note that many students, seeing the above free-body, think the resulting answer of
0.50 should be plotted at the quarter point, but this is not correct. The resulting answer is indeed a shear located at the quarter point, but the result is to be plotted at the current location of the unit load, i.e. at the center of the beam. Likewise, the two answers found when the unit load was positioned at the quarter point of the beam will both be plotted at the quarter point, since the unit load was located there at the time those answers were found – one slightly to the left, and one slightly to the right of the quarter point.
11) Finally, put the unit load at the right end of the beam and solve for the left reaction (see Figure 1.)
12) Again, draw a free-body of the beam seen to the left of the quarter point, and solve for the exposed shear by summing forces vertically (see Figure 10.) In this case, a resulting shear of zero is caused by the unit load placed on the right end of the beam.
V= 0.0
M= 0
Figure 10
13) We now have enough points to plot a continuous curve for the influence line for shear at the quarter point of a simply supported beam:

0.75
0.25
0.50
Figure 11
Influence lines for internal moment in a simply supported beam:
Assume that we now desire to generate the influence line for moment at the quarter point in the same beam. It is found in the same manner – namely, a unit load is moved across the entire length of the beam, and the resulting moment at the quarter point is computed and plotted. The process is detailed below:
1.
Position the unit load at the left reaction and solve for the resulting magnitude of the left reaction (see Figure 1.)
2.
Cut the beam at the quarter point, where the moment M is desired, and draw a free body to expose the internal moment at that point in the beam (see Figure 5.)
3.
Solve for the resulting exposed moment M = 0 using ∑ M about the quarter-point
= 0. In this case a unit load placed at the left end of the beam causes zero moment at the quarter point.
4.
Now move the unit load just slightly to the left of the quarter point on the beam and solve for the resulting reactions (see Figure 6.)
5.
Again, draw a free-body of the beam to the left of the quarter point, and solve for the exposed moment at the quarter-point by summing moments about the quarterpoint (see Figure 7.) In this case, a resulting moment of 0.1875L is caused by a unit load placed just to the left of the quarter-point. The direction of internal moment shown in Figure 7 will arbitrarily be selected as positive.
6.
Now move the unit load just slightly to the right of the quarter point on the beam and solve for the resulting reactions as again seen in Figure 6. Since the load is moved only an infinitesimal distance to the right of its previous position, the reaction remains = 0.75, and even though the unit load now moves outside the free-body used to solve for the resulting moment at the quarter point, the moment remains unchanged (see Figure 8.)
7.
Next, move the unit load to the center of the beam and solve for the left reaction
(see Figure 2.)
8.
Again, draw a free-body of the beam to the left of the quarter point, and solve for the exposed moment by summing moments. In this case, a resulting moment of positive 0.125L is caused at the quarter-point due to the unit load placed at the center of the beam (see Figure 9.)
9.
Note that many students, seeing the above free-body, feel the resulting moment of
0.125L should be plotted at the quarter point, but this is not correct. The resulting

answer is indeed located at the quarter point, but the result is to be plotted at the current location of the unit load, i.e. at the center of the beam. Likewise, the two answers found when the unit load was positioned at the quarter point of the beam will both be plotted at the quarter point, since the unit load was located there at the time those answers were found – one slightly to the left, and one slightly to the right of the quarter point. However, both answers were the same.
10.
Finally, put the unit load at the right end of the beam and solve for the left reaction (see Figure 1.)
11.
Again, draw a free-body of the beam seen to the left of the quarter point, and solve for the exposed moment by summing moments (see Figure 10.) In this case, a resulting moment of zero is caused by the unit load placed on the right end of the beam.
12.
We now have enough points to plot a continuous curve for the influence line for moment at the quarter point of a simply supported beam:
0.1875L
Figure 12
Other Examples for influence lines:
Influence line for shear at the center of a simply supported beam:
0.50
0.50
Figure 13
Influence line for shear at the right quarter of a simply supported beam:

0.25
0.75
Figure 14
Influence line for shear for the right end of a simply supported beam:
1.0
Figure 15
Influence line for moment at the center of a simply supported beam:
0.25L
Figure 16
Influence line for moment at the right quarter of a simply supported beam:

0.1875L
Figure 17
Influence line for moment at 7/8 of L from the left reaction of a simply supported beam:
0.109375L
Figure 18
Teaming Problems:
Use statics to verify the influence line for shear at the quarter point for the doublyoverhanging simple beam shown below. Verify the results at x = 0, L/4, 2L/4, 3L/4, 4L/4,
5L/4, 6L/4, and 7L/4:
Figure 19
Use statics to verify the influence line for moment at the quarter point for the doublyoverhanging simple beam shown below. Verify the results at x = 0, L/4, 2L/4, 3L/4, 4L/4,
5L/4, 6L/4, and 7L/4:

Figure 20